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Prove that every natural number $n\geqslant 12$ can be written as a linear combination of $4$ and $5$ with non negative integer coefficients.

I proceeded by induction.

For $n=12$ we have $12=3\cdot 4 +0\cdot 5$

Let $n>12$

Suppose that the hypothesis is valid for every $k\in \mathbb{N}$ with $12<k<n$

Now by doing some calculations we see that for $n=13,14,\ldots,24$ can be written as a linear combination of $4$ and $5$ with non negative integer coefficients.

Suppose that $n>24$ then $n=12+(12+k)$ where $12<12+k<n$ and by induction hypothesis $12+k=4x+5y$ , $x,y\geqslant0$

We conclude that $n=12+4x+5y$.

Is this proof right or does it have a logical mistake?

My question is a proof verification and about a possible logical mistake in using the strong induction. It's not an exercise that I want someone to solve for me.

Larry B.
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Marios Gretsas
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  • Your idea is correct. Your formalization is flawed. – b00n heT Oct 21 '16 at 21:02
  • It's not a mistake, but if you want to present this as a proof by induction, the part where you check the result from 13 to 24 should be part of the base case. You don't need that much by the way : if it works for $12$, $13$, $14$ and $15$ then it works for all numbers $\geq 12$. – Arnaud D. Oct 21 '16 at 21:05
  • so my use of induction is wrong – Marios Gretsas Oct 21 '16 at 21:06

2 Answers2

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The logical mistake is that you do not prove the $k+1$ case. You are also mixing $n$ and $k$ in a way that does not make sense for an induction proof. I think you are essentially going down the path of Larry B.'s answer but try to shoehorn it into an induction formalism. If you are insisting on induction you can try redefining your base case to be all numbers from 12 to 24, and proceed from there, but I think it will become unwieldy.

Since you asked in the comments to the accepted answer, here's my take on how I would prove this by induction:

Base case

For $n=12$ the statement is true ($3\cdot4+0\cdot5=12$)

Induction Hypothesis

Assume the statement is true for $k\ge12$

Induction step

If the statement is true for $k\ge12$ prove that it is true for $k+1$

By the induction hypothesis $k =x\cdot4+y\cdot5$, with $x,y\ge0$

  • If $x>0$ then

$$\begin{align} k+1 &= x\cdot4+y\cdot5 + 1 \\ &= (x-1)\cdot4 + y\cdot5 +4 +1 \\ &= (x-1)\cdot4 + y\cdot5 +5 \\ &= (x-1)\cdot4 + (y+1)\cdot5\\ \end{align}$$

  • If $x = 0$ then $k= y\cdot5$, with $\mathbf{y\ge3}$, since $n\ge12$. Hence $$\begin{align} k+1 &= y\cdot5 +1 \\ &= (y-3)\cdot5 +15+1 \\ &= (y-3)\cdot5 +16 \\ &= (y-3)\cdot5 + 4\cdot4\\ \end{align}$$

So we have proven that $k+1$ can be written in the form of $x'\cdot4+y'\cdot5$, with $x',y'\ge0$

Thanassis
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    This is an elegant proof. Not only does it use induction, as required by OP, but unlike my solution it does not require any more manual computation than necessary. @capo, you should consider switching to this as your answer, if you can. – Larry B. Oct 22 '16 at 04:14
  • it's indeed a very nice elegant solution..good job..thank you! – Marios Gretsas Oct 22 '16 at 11:42
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It's the right idea, but you can do it without induction.

We'll take your idea of doing the first dozen computations, so we need to prove for $k > 24$. Let $0\leq a<12$ be such that $a \equiv k \mod 12$, so $k = (12 + a) + 12\cdot j$. We know that $12\cdot j = 4\cdot (3 \cdot j)$, and we have the manual computations for $12 + a = 4\cdot x + 5 \cdot y$, thus $$k = 4\cdot (x + j\cdot 3) + 5\cdot y$$

Larry B.
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